Chapter 3Chapter 3

Molecules, Ions, & their Molecules, Ions, & their CompoundsCompounds

Chapter goalsChapter goals

• Interpret, predict, and write formulas for Interpret, predict, and write formulas for ionic and molecular compounds.ionic and molecular compounds.

• Name compounds.Name compounds.• Understand some properties of ionic Understand some properties of ionic

compounds.compounds.• Calculate and use molar mass.Calculate and use molar mass.• Calculate percent composition for a Calculate percent composition for a

compound and derive formulas from compound and derive formulas from experimental data.experimental data.

MoleculeMolecule

• an assembly of 2 or more atoms (mostly an assembly of 2 or more atoms (mostly of non-metals) bound together in a of non-metals) bound together in a particular ratio and a particular mannerparticular ratio and a particular manner

• is the smallest identifiable unit into is the smallest identifiable unit into which a pure which a pure substancesubstance can be divided can be divided and still retain the composition and and still retain the composition and chemical properties of the substancechemical properties of the substance

Elements that exist as Elements that exist as moleculesmolecules

• Atoms of most of the nonmetals form Atoms of most of the nonmetals form discrete molecules, except for the noble discrete molecules, except for the noble gases.gases.

• Some elements can exist in more than one Some elements can exist in more than one form of molecule; the different forms are form of molecule; the different forms are called called allotropesallotropes. Examples are:. Examples are:

• Diamond, graphite, and buckyballs for Diamond, graphite, and buckyballs for carbon.carbon.

• OO2 2 and Oand O33 (ozone) for oxygen. (ozone) for oxygen.

ELEMENTS THAT ELEMENTS THAT EXIST AS MOLECULESEXIST AS MOLECULES

ELEMENTS THAT ELEMENTS THAT EXIST AS MOLECULESEXIST AS MOLECULES

Allotropes of CAllotropes of CAllotropes of CAllotropes of C

ELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS POLYATOMIC MOLECULESPOLYATOMIC MOLECULESELEMENTS THAT EXIST AS ELEMENTS THAT EXIST AS POLYATOMIC MOLECULESPOLYATOMIC MOLECULES

White P4 and polymeric red phosphorus

Sulfur: crown-shaped rings of S8 molecules

Molecular compoundsMolecular compoundsmodels - shapesmodels - shapes

Methane, CH4Water, H2O

Ammonia, NH3

Molecular compoundsMolecular compounds

• Are made of non-metalsAre made of non-metals

• HH22O, carbon dioxide (COO, carbon dioxide (CO22), ammonia (NH), ammonia (NH33), ),

nitric acid (HNOnitric acid (HNO33), ethanol (CH), ethanol (CH33CHCH22OH), OH),

sulfuric acid (Hsulfuric acid (H22SOSO44), glucose (C), glucose (C66HH1212OO66), are ), are

examples among thousandsexamples among thousands

• In the molecules of theses compounds, In the molecules of theses compounds, atoms atoms share pairs of electrons pairs of electrons

Molecular compounds: FormulasMolecular compounds: FormulasNAME MOLECULAR CONDENSED STRUCTURAL FORMULANAME MOLECULAR CONDENSED STRUCTURAL FORMULA FORMULAFORMULA FORMULA FORMULA

H H | |Ethanol CEthanol C22HH66O O CH3CH2OH H─C ─ C─O─H

| | H H

H H | |DimethylDimethyl C C22HH66O O CH3OCH3 H─C ─ O─ C─H

etherether | | H H

Ethanol and dimethyl ether are said to be structural Ethanol and dimethyl ether are said to be structural isomers.

Ionic CompoundsIonic Compounds

Ion Ion • charged particle (atom or group of charged particle (atom or group of

atoms)atoms)

• cation: + chargecation: + charge

• anion: – chargeanion: – charge

Ionic CompoundsIonic Compounds

Sodium chloride or Sodium chloride or ““table salttable salt”” is an is anexample of an ionic compound.example of an ionic compound.

Ionic compoundsIonic compounds

• consist of positive and negative ions, consist of positive and negative ions, mostly a metal and a non-metal, mostly a metal and a non-metal, respectively.respectively.

• have attractions called have attractions called ionic bondsionic bonds between positively (between positively (cationscations) and ) and negatively charged ions (negatively charged ions (anionsanions). ).

• have high melting and boiling points.have high melting and boiling points. TTmm of NaCl = 800 °C = 1472 °F of NaCl = 800 °C = 1472 °F• are solid at room temperature.are solid at room temperature.

Charge BalanceCharge Balance

Group IA VIIA NaCl, sodium chloride IIA of periodic table

Periodic TablePeriodic Table

Monatomic CationsMonatomic CationsMetal atoms of group 1A Metal atoms of group 1A lose one electron to lose one electron to produce a mono-positive ionproduce a mono-positive ion

Monatomic CationsMonatomic CationsMetal atoms of group 2A Metal atoms of group 2A lose two electrons to lose two electrons to produce a di-positive ionproduce a di-positive ion

Monatomic AnionsMonatomic AnionsNonmetals often gain one or more electrons andNonmetals often gain one or more electrons andform ions having a negative charge equal toform ions having a negative charge equal to the group numberthe group number of the element of the element minus 8minus 8::

Group Atom gained eGroup Atom gained e-- Resulting Resulting anion anion 5A N 5A N 3 3 NN3-3- PP 6A O 6A O 2 2 O O2-2- SS

7A F 7A F 1 1 F F-- Cl, Br, I ClCl, Br, I Cl−− , Br , Br−− , I , I−−

Charges of Representative ElementsCharges of Representative Elements

Monatomic CationsMonatomic CationsTransition metals (B-group elements) can formTransition metals (B-group elements) can forma no easily predictable variety of cations:a no easily predictable variety of cations:Group Atom Electrons loss ResultingGroup Atom Electrons loss Resulting cation cation 7B Mn 7B Mn 2 2 MnMn2+2+ 8B Fe 8B Fe 2 2 FeFe2+2+ 8B Fe 8B Fe 3 3 FeFe3+3+ 1B Cu 1B Cu 1 1 CuCu++ 1B Cu 1B Cu 2 2 CuCu2+2+ 2B Zn 2B Zn 2 2 ZnZn2+2+ 2B Cd 2B Cd 2 2 CdCd2+2+

Transition Metals form Positive IonsTransition Metals form Positive Ions

Most transition metals and Group 4A metals form 2 or more positive ions.

Zn, Ag, and Cd form only one ion.

Names of Some Common IonsNames of Some Common IonsMain group metals Nonmetal: change the lastMain group metals Nonmetal: change the lastelement name onlyelement name only part of name to part of name to ideide

Naming CationsNaming Cations• transition metals and In, Sn, Tl, Pb, Bitransition metals and In, Sn, Tl, Pb, Bi

new system: new system: element name (charge in element name (charge in Roman numerals)Roman numerals)

eg. Mneg. Mn2+2+ manganese(II) manganese(II)

MnMn3+3+ manganese(III) manganese(III)

CrCr2+2+ chromium(II) Cr chromium(II) Cr3+3+ chromium(III) chromium(III)

FeFe2+2+ iron(II) Fe iron(II) Fe3+3+ iron(III) iron(III)

Exceptions: when only one cationExceptions: when only one cation

AgAg+ + silver Zn silver Zn2+2+ zinc Cd zinc Cd2+2+ cadmium cadmium

old systemold system

Latin name-suffixLatin name-suffix

suffix = -ic for higher charge, -ous for suffix = -ic for higher charge, -ous for lower chargelower charge

eg. Cueg. Cu++ copper(I) copper(I) cuprous ioncuprous ion

CuCu2+2+ copper(II) copper(II) cupric ioncupric ion

CoCo2+2+ cobalt(II) cobalt(II) cobaltous ioncobaltous ion

CoCo3+3+ cobalt(III) cobalt(III) cobaltic ioncobaltic ion

FeFe2+2+ iron(II) iron(II) ferrous ionferrous ion

FeFe3+3+ iron(III) iron(III) ferric ionferric ion

Polyatomic IonsPolyatomic IonsA A polyatomic ion

• is a group of atoms. is a group of atoms.

• has an overall ionic charge, positive or has an overall ionic charge, positive or negative.negative.

Polyatomic Ions (memorize)Polyatomic Ions (memorize)Some examples of polyatomic ions areSome examples of polyatomic ions are

NHNH44++ ammoniumammonium HH33OO++ hydroniumhydronium

OHOH−− hydroxide hydroxide NN33−− azideazide

COCO3322−− carbonatecarbonate CNCN−− cyanidecyanide

CHCH33COCO22−− acetate acetate CC22OO44

22−− oxalate oxalate

NONO33−−

nitratenitrate NONO22−−

nitritenitrite

POPO4433−− phosphatephosphate POPO33

33−− phosphitephosphite

SOSO4422−− sulfatesulfate SOSO33

22−− sulfitesulfite

CrOCrO4422−− chromatechromate CrCr22OO77

22−− dichromatedichromate

MnOMnO44−− permanganatepermanganate MnO MnO44

2− 2− manganatemanganate

Hydrogenated Polyatomic IonsHydrogenated Polyatomic Ions

HCOHCO33−− hydrogen carbonate (bicarbonate)hydrogen carbonate (bicarbonate)

HSOHSO44−− hydrogen sulfate (bisulfate)hydrogen sulfate (bisulfate)

HSOHSO33−− hydrogen sulfite (bisulfite)hydrogen sulfite (bisulfite)

HPOHPO442−2− hydrogen phosphatehydrogen phosphate

HH22POPO44−− dihydrogen phosphatedihydrogen phosphate

HSHS−− hydrogen sulf hydrogen sulfide (from sulfide, Side (from sulfide, S22−−))

SystematicsSystematics• ClOClO44

– – perchlorateperchlorate

• ClOClO33– – chloratechlorate

• ClOClO22– – chloritechlorite

• ClOClO– – hypochloritehypochlorite

The same with other halogens (except for The same with other halogens (except for F)F)

• IOIO44–– periodateperiodate

• IOIO33–– iodateiodate

• IOIO22–– ioditeiodite

• IOIO–– hypoioditehypoiodite

1

Cr98

H

Li

Na

K

Rb

Cs

Fr

3

11

19

37

55

87

1.008

6.941

22.99

39.10

85.47

132.9

4

12

20

38

56

88

9.012

24.31

40.08

87.62

137.3

Be

Mg

Ca

Sr

Ba

Ra

21

39

57

89

44.96

88.91

138.9

Sc

Y

La

Ac

22

40

72

104

47.88

91.22

178.5

Ti

Zr

Hf

Unq

23

41

73

105

50.94

92.91

180.9

V

Nb

Ta

Unp

24

42

74

106

52.00

95.94

183.8

Mo

W

Unh

25

43

75

54.94

186.2

Mn

Tc

Re

Fe

Ru

Os

26

44

76

55.85

101.1

190.2

27 28 29 30

58.93 58.69 63.55 65.39Co Ni Cu Zn

Rh Pd Ag Cd

Ir Pt Au Hg

45 46 47 48

77 78 79 80102.9 106.4 107.9 112.4

192.2 195.1 200.6197.0

2

5 6 7 8 9 10

13 14 15 16 17 18

31 32 33 34 35 36

49 50 51 52 53 54

81 82 83 84 85 86

4.003

10.81 12.01 14.01 16.00 19.00 20.18

26.98 28.09 30.97 32.07 35.45 39.95

69.72 72.59 74.92 78.96 79.90 83.80

114.8 118.7 121.8 127.6 126.9 131.3

204.4 207.2 209.0

He

B C N O F Ne

Al Si P S Cl Ar

Ga Ge As Se Br Kr

In Sn Sb Te I Xe

Tl Pb Bi Po At Rn

Cr

1

1A

2A

3B 4B 5B 6B 7B 8B 1B 2B

3A 4A 5A 6A 7A

8A

58 59 60 61 62 63 64 65 66 67 68 69 70 71

90 91 92 93 94 95 96 97 98 99 100 101 102 103140.1 140.9 144.2 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0

232.0 238.0

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Fm Md No LwTh Pa U Np Pu Am Cm Bk Cf Es

(223) (226) (227) (257) (260) (263)

(210) (210) (222)

(257)(254)(256)(253)(254)(249)(247)(243)(242)(237) (247)(231)

(147)

(98)

107 108 109Uns Uno Une(262) (265) (266)

1

2

3 4 5 6 7 8 9 10 11 12

13 14 15 16 17

18

8B8B

• POPO553– 3– perphosphateperphosphate

• POPO443– 3– phosphatephosphate

• POPO333– 3– phosphitephosphite

• POPO223– 3– hypophosphitehypophosphite

• AsOAsO553–3– perarsenateperarsenate

• AsOAsO443–3– arsenatearsenate

• AsOAsO333–3– arsenitearsenite

• AsOAsO223–3– hypoarsenitehypoarsenite

Oxyanions that end in Oxyanions that end in ateate

Group Group

III AIII A

Group Group

IV AIV A

Group Group

V AV A

Group Group

VI AVI A

Group Group

VII AVII A

BOBO333–3–

borateborate

COCO332–2–

carbonatecarbonate

NONO33––

nitratenitrate

SiOSiO332–2–

silicatesilicate

POPO443–3–

phosphatephosphate

SOSO442–2–

sulfatesulfate

ClOClO33––

chloratechlorate

AsOAsO443–3–

arsenatearsenate

SeOSeO442–2–

selenateselenate

BrOBrO33––

bromatebromate

TeOTeO442–2–

telluratetellurate

IOIO33––

iodateiodate

Other oxyanions:Other oxyanions: Once ions ending in Once ions ending in ate are memorized, those ate are memorized, those with different number of O atoms are named with with different number of O atoms are named with

following following prefixesprefixes and and suffixessuffixes 2 oxygen less 2 oxygen less than the than the ““ateate”” ion ision is

hypo__ite hypo__ite

1 oxygen less 1 oxygen less than the than the ““ateate”” ion ision is

___ite___ite

““ateate”” ion is ion is

____ate____ate

1 oxygen 1 oxygen more than more than ““ateate”” ion is ion is

per__ateper__ate

BrOBrO–– hypobromite hypobromite BrOBrO22

–– bromitebromite

BrOBrO33–– bromatebromate

BrOBrO44–– perbromateperbromate

Formula UnitFormula Unit• combination of ions in simplest whole combination of ions in simplest whole

number ratio to be electrically neutral number ratio to be electrically neutral (the (the simplest unit of an ionic compound)simplest unit of an ionic compound)

• consists of positively and negatively charged consists of positively and negatively charged ions.ions.

• is neutral.is neutral.

• has charge balance.has charge balance. total positive charge = total negative chargetotal positive charge = total negative charge

The symbol of the metal is written first The symbol of the metal is written first followed by the symbol of the nonmetal.followed by the symbol of the nonmetal.

Naming Ionic Compounds with Naming Ionic Compounds with Two Elements (Two Elements (binary compoundbinary compound))

To name a compoundTo name a compound that contains twothat contains two elements,elements,

• identify theidentify the cationcation and and anion.anion.

• name the cation name the cation first followed by the first followed by the name of the anion.name of the anion.

Examples, name each belowExamples, name each below

• KBrKBr

KK++

potassiumpotassium

BrBr––

bromidebromide

potassium bromidepotassium bromide

• AlFAlF33

AlAl3+3+

aluminumaluminum

FF– –

fluoridefluoride

aluminum fluoridealuminum fluoride

• SrSr33PP2 2 Sr P Sr P

SrSr2+2+ 3 3(+2) = +6 2(+2) = +6 2((−3) = −3) = −6−6

strontiumstrontium

PP3–3–

phosphidephosphide

strontium phosphidestrontium phosphide

• CuClCuCl22 Cu Cl Cu Cl ClCl–– 1 1(+2) = +2 2(+2) = +2 2((−1) = −2−1) = −2 chloridechloride CuCu2+2+

copper(II)copper(II) cupriccupric copper(II) chloridecopper(II) chloride cupric chloridecupric chloride

CuCl: copper(I) chlorideCuCl: copper(I) chloride

• WFWF66 W F W F

FF–– w + 6x( w + 6x(−1) = 0−1) = 0

fluoride w = 6+ fluoride w = 6+

WW6+6+

tungsten(VI)tungsten(VI)

tungsten(VI) fluoridetungsten(VI) fluoride

Naming Compounds with Naming Compounds with Polyatomic IonsPolyatomic Ions

The positive ion is named first followed by theThe positive ion is named first followed by the

name of the polyatomic ion.name of the polyatomic ion.

NaNaNONO33 sodiumsodium nitratenitrate

KK22SOSO44 potassiumpotassium sulfatesulfate

FeFe(HCO(HCO33))33 iron(III)iron(III) bicarbonatebicarbonate

or iron(III)or iron(III) hydrogen carbonatehydrogen carbonate

fe + 3x(fe + 3x(−1) = 0 fe = 3+ iron(III)−1) = 0 fe = 3+ iron(III)

(NH(NH44))33POPO33 ammoniumammonium phosphitephosphite

Writing Formulas with Polyatomic IonsWriting Formulas with Polyatomic IonsThe formula of an ionic compound The formula of an ionic compound

• containing a polyatomic ion must have a containing a polyatomic ion must have a charge balance that equals zero (0).charge balance that equals zero (0).

NaNa++ and and NONO33−− NaNaNONO33

• with two or more polyatomic ions has the with two or more polyatomic ions has the polyatomic ions in parentheses.polyatomic ions in parentheses.

MgMg2+2+ and and 2NO2NO33−−

Mg Mg(NO(NO33))22

subscript subscript 22 for charge balance for charge balance

• Aluminum sulfateAluminum sulfate

2Al2Al3+3+ and 3SO and 3SO4422−− Al Al22(SO(SO44))33

Learning CheckLearning CheckMatch each formula with the correct name.Match each formula with the correct name.

A. MgSA. MgS 1) magnesium sulfite 1) magnesium sulfite

MgSOMgSO33 2) magnesium sulfate2) magnesium sulfate

MgSOMgSO44 3) magnesium sulfide3) magnesium sulfide

B. Ca(ClOB. Ca(ClO33))22 1) calcium chlorate1) calcium chlorate

CaClCaCl22 2) calcium chlorite2) calcium chlorite

Ca(ClOCa(ClO22))2 2 3) calcium chloride3) calcium chloride

Name each of the following compounds:Name each of the following compounds:

A.A. Mg(NOMg(NO33))22 magnesium nitrate Mgmagnesium nitrate Mg2+2+ and 2 NO and 2 NO33−−

B.B. Cu(ClOCu(ClO33))22 copper(II) chlorate Cucopper(II) chlorate Cu2+2+ and 2 ClO and 2 ClO33−−

C.C. PbOPbO22 lead(IV) oxide Pblead(IV) oxide Pb4+4+ and 2 O and 2 O22−−

D.D. FeFe22(SO(SO44))33 iron(III) sulfate 2 Feiron(III) sulfate 2 Fe3+3+ and 3 SO and 3 SO442 −2 −

E.E. BaBa33(PO(PO33))22 barium phosphite 3 Babarium phosphite 3 Ba2+2+ and 2 PO and 2 PO333 −3 −

Learning CheckLearning CheckSelect the correct formula for each.Select the correct formula for each.

A. A. aluminum nitratealuminum nitrate

1) AlNO1) AlNO3 3 2) Al(NO)2) Al(NO)33 3) Al(NO3) Al(NO33))33

B. B. copper(II) nitratecopper(II) nitrate

1) CuNO1) CuNO33 2) Cu(NO2) Cu(NO33))22 3) Cu3) Cu22(NO(NO33))

C. iron(III) hydroxideC. iron(III) hydroxide

1) FeOH1) FeOH 2) Fe2) Fe33OHOH 3) Fe(OH)3) Fe(OH)33

D. D. tin(IV) hydroxidetin(IV) hydroxide

1) Sn(OH)1) Sn(OH)44 2) Sn(OH)2) Sn(OH)22 3) Sn3) Sn44(OH)(OH)

• CaSOCaSO44

CaCa2+2+ 1 1(+2) = +2 1(+2) = +2 1((−2) = −2−2) = −2

calciumcalcium

SOSO442–2–

sulfatesulfate

calcium sulfatecalcium sulfate

• (NH(NH44))22SS

NHNH44++

ammoniumammonium

SS2–2–

sulfidesulfide

ammonium sulfideammonium sulfide

• (NH(NH44))33POPO44

NHNH44++

ammoniumammonium

POPO443–3–

phosphatephosphate

ammonium phosphateammonium phosphate

• Mo(BrO)Mo(BrO)66

BrOBrO–– mo + mo + 66x(x(−1) = 0 mo = 6+−1) = 0 mo = 6+

hypobromitehypobromite

MoMo6+6+

molybdenum(VI)molybdenum(VI)

molybdenum(VI) hypobromitemolybdenum(VI) hypobromite

Write formula forWrite formula for

• rubidium bromiderubidium bromide

RbRb

RbRb++

BrBr

BrBr– –

RbBrRbBr

• calcium phosphidecalcium phosphide

CaCa

CaCa2+2+

P For neutrality we need P For neutrality we need

PP3–3– 3 Ca 3 Ca2+2+ and 2 P and 2 P3–3–

CaCa33PP22 that is, that is, +6+6 – 6 = 0 – 6 = 0

• niobium(IV) sulfiteniobium(IV) sulfite

NbNb4+ 4+ In order to have the sameIn order to have the same

SOSO332–2– total + and – charge, we needtotal + and – charge, we need

one Nbone Nb4+4+ ion and two SO ion and two SO332–2–, that is , that is

11(+4) + 2 (+4) + 2 (–2) = 4 – 4 =0 (neutrality)(–2) = 4 – 4 =0 (neutrality)

Nb(SONb(SO33))22

• barium phosphite: Babarium phosphite: Ba2+2+ and PO and PO3333––

for the compound to be neutralfor the compound to be neutral, we need, we need

3 Ba3 Ba2+2+ and 2 PO and 2 PO333–3–, that is, , that is, +6+6 – 6 = 0 – 6 = 0

BaBa33(PO(PO33))22

Electrostatic ForcesElectrostatic Forces

The oppositely charged ions in ionicThe oppositely charged ions in ionic

compounds are attracted to one another bycompounds are attracted to one another by ELECTROSTATIC FORCESELECTROSTATIC FORCES..

These forces are governed byThese forces are governed by COULOMBCOULOMB’’S S LAWLAW..

Electrostatic ForcesElectrostatic Forces

COULOMBCOULOMB’’S LAWS LAW

As ion charge As ion charge increasesincreases, the attractive force , the attractive force increasesincreases..

As the distance between ions (d) As the distance between ions (d) increasesincreases, the , the attractive force attractive force decreases decreases..

This idea is important and will come up many This idea is important and will come up many times in future discussions!times in future discussions!

F F QQccQQaa

dd22

NaCl NaNaCl Na++ Cl Cl––

CaClCaCl22 Ca Ca2+2+ Cl Cl––

CaS CaCaS Ca2+2+ S S–2–2

AlAl22SS33 Al Al3+3+ S S2–2–

Product of charges

1

6 strongest force

2

4

F F QQccQQaa

dd22

FF––NaNa++ NaNa++

NaNa++NaNa++

ClCl––

BrBr––II––

dddd

dd

dd

NaF has the strongest force and NaI the weakest

AcidsAcids

• HH++ is only cation is only cation in acidsin acids

Binary AcidsBinary Acids hydrogen bonded to one other hydrogen bonded to one other

elementelement

hydro-anion-ic acidhydro-anion-ic acid

ExamplesExamples• HCl HCl in aqueous solution (water is the solvent)in aqueous solution (water is the solvent) HH++ and Cl and Cl– –

anion is chlorideanion is chloride hydro-chlor-ic acid hydro-chlor-ic acid Same for F, Br, and ISame for F, Br, and I hydrohydrochlorchloricic acid acid

• HH22SS hydrosulfuric acidhydrosulfuric acid

• HH33PP hydrophosphoric acidhydrophosphoric acid

• HH22TeTe hydrotelluric acidhydrotelluric acid

OxoacidsOxoacids hydrogen, oxygen, and another hydrogen, oxygen, and another

elementelement

anion-suffix acidanion-suffix acid

anionanion

suffix: -ate suffix: -ate -ic, -ite -ic, -ite -ous -ous

acidacid

Examples, nameExamples, name

• HH33POPO44 HH22SOSO44

phosphoric acidphosphoric acid sulfuric acidsulfuric acid POPO44

3–3– SOSO442–2–

phosphatephosphate sulfatesulfate

• HH33POPO33 HH22SOSO33

phosphorous acidphosphorous acid sulfurous acidsulfurous acid POPO33

3–3– SOSO332–2–

phosphitephosphite sulfitesulfite

If more than two oxoacids, that with less OIf more than two oxoacids, that with less Othan the –ous will be hypo– –ous; that with than the –ous will be hypo– –ous; that with more O than the –ic will be per– –ic. more O than the –ic will be per– –ic.

• HBrOHBrO HBrOHBrO22

hypohypobromous acidbromous acid bromous acidbromous acid BrOBrO–– BrOBrO22

––

hypobromite hypobromite bromitebromite

• HBrOHBrO33 HBrOHBrO44

bromic acidbromic acid perperbromic acidbromic acid BrOBrO33

–– BrOBrO44––

bromatebromate perbromateperbromate bromic acidbromic acid perperbromic acidbromic acidThe same applies to Cl and I,The same applies to Cl and I, not to Fnot to F

Molecular CompoundsMolecular Compounds

• contain only nonmetals bound by covalent bonds

methane, CH4

Covalent bonds form

• when atoms share electrons to complete octets. There are no ions.

Naming Molecular (Covalent) CompoundsNaming Molecular (Covalent) Compounds

To name covalent To name covalent compoundscompounds• STEP 1STEP 1:: Name the first Name the first

nonmetal as an nonmetal as an element. element.

• STEP 2STEP 2:: Name the Name the second nonmetal with second nonmetal with an an ide ending. ending.

• STEP 3STEP 3:: Use Use prefixes to to indicate the number of indicate the number of atoms (atoms (subscriptsubscript) of ) of each element.each element.

Prefixes Used in Naming Covalent Compounds # of Atoms Prefix

1 Mono2 Di3 Tri4 Tetra5 Penta6 Hexa7 Hepta8 Octa9 Nona

10 Deca

ExceptionException

• prefix mono- omitted from name of prefix mono- omitted from name of first elementfirst element

• eg. COeg. CO22

carbon dioxide, carbon dioxide, not monocarbon not monocarbon dioxidedioxide

Examples, nameExamples, nameWhat is the name of SOWhat is the name of SO33??

1. The first nonmetal is S sulfur. 1. The first nonmetal is S sulfur.

2. The second nonmetal is O named ox2. The second nonmetal is O named oxideide..

3.3. The subscript 3 of O is shown as theThe subscript 3 of O is shown as the

prefix prefix tri..

SOSO3 3 sulfur sulfur trioxoxideide

The subscript 1 (for S) or The subscript 1 (for S) or monomono is is understoodunderstood

Examples, nameExamples, name

• PP22OO55

phosphorusphosphorus

Diphosphorus (Diphosphorus (two P atomstwo P atoms))

oxygenoxygen

oxideoxide

Pentoxide(Pentoxide(5 O atoms5 O atoms) (not pent) (not pentaoaoxide: a xide: a of penta- is dropped)of penta- is dropped)

diphosphorus pentoxidediphosphorus pentoxide

• PP33BrBr66

phosphorusphosphorus

triphosphorustriphosphorus

brominebromine

bromidebromide

hexabromidehexabromide

triphosphorus hexabromidetriphosphorus hexabromide

Learning CheckSelect the correct name for each compound.Select the correct name for each compound.

A.A. SiClSiCl44 1) silicon chloride1) silicon chloride

2) tetrasilicon chloride2) tetrasilicon chloride3) silicon tetrachloride3) silicon tetrachloride

B. B. PP22OO55 1) phosphorus oxide1) phosphorus oxide

2) phosphorus pentoxide2) phosphorus pentoxide3) diphosphorus pentoxide3) diphosphorus pentoxide

C.C. ClCl22OO77 1) dichlorine heptoxide1) dichlorine heptoxide

2) dichlorine oxide2) dichlorine oxide3) chlorine heptoxide 3) chlorine heptoxide

Learning Check

Write the name of each molecular (covalent)Write the name of each molecular (covalent)

compound.compound.

COCO __________________________________________

COCO22 __________________________________________

PClPCl33 __________________________________________

CClCCl44 __________________________________________

NN22OO __________________________________________

Learning Check

Write the correct formula for each of theWrite the correct formula for each of the

following.following.

A. phosphorus pentachlorideA. phosphorus pentachloride

B. dinitrogen trioxideB. dinitrogen trioxide

C. sulfur hexafluorideC. sulfur hexafluoride

Common NamesCommon Names

• HH22O O waterwater

• PHPH33 phosphinephosphine

• CaCOCaCO33 limestonelimestone

• CaO CaO limelime

• Ca(OH)Ca(OH)22 slaked limeslaked lime

• CHCH44 methanemethane

• NaCl NaCl table salttable salt

• NN22O O laughing gaslaughing gas

• NaHCONaHCO33 baking sodabaking soda

• NaNa22COCO33•10H•10H22O O washing sodawashing soda

Sodium carbonate Sodium carbonate decadecahydratehydrate

• MgSOMgSO44•7H•7H22O O epsom saltepsom salt

• Mg(OH)Mg(OH)22 milk of magnesiamilk of magnesia

• Ca(SOCa(SO44)•2H)•2H22O O gypsumgypsum

Nomenclature DetailsNomenclature Details• acid (acid (aqueous solutionaqueous solution))

HBrHBr(g) (g) hydrogen bromide (hydrogen bromide (gas phasegas phase))

HBrHBr(aq) (aq) hydrobromic acid (hydrobromic acid (aqueous aqueous

solutionsolution))

Same for HF, HCl, and HI.Same for HF, HCl, and HI.

HydrogenHydrogen• belongs in group all its ownbelongs in group all its own

• generally considered a nonmetalgenerally considered a nonmetal

• forms both Hforms both H++ and H and H–– in ionic compounds in ionic compounds

HCl (in aqueous solution)HCl (in aqueous solution)

HH++ and Cl and Cl––

NaH (sodium hydride, same with Li, K, NaH (sodium hydride, same with Li, K, Rb, Ca, Ba…) Rb, Ca, Ba…)

NaNa++ and and HH––

• mostly forms molecular compoundsmostly forms molecular compounds

e.g.. CHe.g.. CH44, NH, NH33, C, C22HH66SO, ...SO, ...

Chemical FormulasChemical Formulas• empirical formulaempirical formula

– indicates the elements present and indicates the elements present and their their simplest, whole-numbersimplest, whole-number ratio in a ratio in a compoundcompound

• molecular formulamolecular formula– indicates elements present and the indicates elements present and the

exact number of atoms of each in a exact number of atoms of each in a unit of a compoundunit of a compound

• structural formulastructural formula– a molecular formula that includes a molecular formula that includes

structural informationstructural information

BenzeneBenzene

structural formulastructural formula

molecular formula Cmolecular formula C66HH66

dividing by smallest subscript (6)dividing by smallest subscript (6)empirical formula CH empirical formula CH

CCCC CC

CC CCCC

HH

HH

HH

HH

HH

HH

DimethyletherDimethylether

• condensed formula Hcondensed formula H33COCHCOCH33

• molecular formula Cmolecular formula C22HH66O (O (cannot be divided cannot be divided

to get whole numbers; by diving by 2 we get to get whole numbers; by diving by 2 we get OO0.50.5))

• empirical formula Cempirical formula C22HH66OO

H H | |DimethylDimethyl C C22HH66O O CH3OCH3 H─C ─ O─ C─H

etherether | | H H

Calcium Chloride (Calcium Chloride (ionicionic))

• structural formula structural formula none (it is ionic)none (it is ionic)

• molecular formula molecular formula none (it is ionic)none (it is ionic)

• formula unitformula unit CaCl CaCl22

• empirical formula CaClempirical formula CaCl22

Other examplesOther examplesNAME Molecular EmpiricalNAME Molecular Empirical

formula formulaformula formula

butanoic acid Cbutanoic acid C44HH88OO22 C C44HH88OO22 CC22HH44OO 2 2 2 22 2 diboron diboron hexahydride hexahydride BB22HH66 B B22HH66 BH BH33

22 22

sodium ditionatesodium ditionate NaNa22SS22OO44 NaSO NaSO22

hexane hexane CC66HH14 14 CHCH2.3332.333 C C33HH77

Molar Mass of a CompoundMolar Mass of a Compoundis the mass (g) of one mole (6.022 x 1023) of• formula units for ionic compounds• molecules for molecular compoundsThe molar massThe molar mass of a compound is the sum of of a compound is the sum of

the molar masses of the elements in the the molar masses of the elements in the

Formula (Formula (times respective coefficientstimes respective coefficients.) .)

• molecular molecular or formulaor formula weight = weight = amu/moleculeamu/molecule• molar mass = g/mol of molecules molar mass = g/mol of molecules = g/mol of= g/mol of formula units (for formula units (for ionic compounds)ionic compounds)

Molar Mass of a CompoundMolar Mass of a CompoundExampleExample: Calculate the molar mass of CaCl: Calculate the molar mass of CaCl22 (ionic) (ionic)

Element Number of Moles Atomic Mass Total Mass

Ca 1 40.1 g/mole 40.1 g

Cl 2 35.5 g/mole 71.0 g

CaClCaCl22 111.1 g/mole111.1 g/mole

For glucose: C6H12O6 (molecular, covalent)

C 6 12.0 g/mole 72.0 g

H 12 1.0 g/mole 12.0 g

O 6 16.0 g/mole 96.0 g C6H12O6 180.0 g/mol

Molar Mass (Molar Mass (MM), Avogadro), Avogadro’’s number, s number, and atoms of elementsand atoms of elements

1 mole = 6.022 x 1023 particles = molar mass(g) molecules or formula units for compounds

From the molecular formula of glucose, C6H12O6,

1 glucose molecule contains 6 C, 12 H, and 6 O atoms

1 mol glucose contains 6 mol C,12 mol H, and 6mol O

x 6.022 x 1023 particles (molecules and atoms)M (g/mol) = 6x AW C + 12 AW H + 6 AW O (from periodic table)

Molar Mass and AvogadroMolar Mass and Avogadro’’s numbers numberAvogadroAvogadro’’s number (s number (6.022 x 106.022 x 102323) can be written ) can be written as equalities and conversion factors.as equalities and conversion factors.

Equality: Equality: (atoms, molecules, ions, electrons, protons)(atoms, molecules, ions, electrons, protons)1 mole = 6.022 x 101 mole = 6.022 x 102323 particles = molar mass (g) particles = molar mass (g)Particles here are molecules or Formula Units (ionic)

6.022 x 106.022 x 102323 particles particles and and 1 mole 1 mole 1 mole1 mole 6.022 x 10 6.022 x 102323 particles particles

6.022 x 106.022 x 102323 particles particles and and molar mass (g) molar mass (g) molar mass (g) molar mass (g) 6.022 x 10 6.022 x 102323 particles particles

1 mole 1 mole and and molar mass (g) molar mass (g) molar mass (g) molar mass (g) 1 mole 1 mole

The mass (in grams) of a single The mass (in grams) of a single moleculemolecule

The molar mass of water, HThe molar mass of water, H22O, is O, is

2 x 1.008 + 16.00 = 18.02 g/mol 2 x 1.008 + 16.00 = 18.02 g/mol (about 18 mL)(about 18 mL)

Does a Does a water molecule have a masswater molecule have a mass of 18.02 g? of 18.02 g?

No, it doesnNo, it doesn’’t.t. 18.02 g (the molar mass)18.02 g (the molar mass)

1 mol H1 mol H22O O —————————— 6.022 x 106.022 x 102323 H H22O moleculesO molecules

by diving, we get by diving, we get 2.99x102.99x10−23−23 g per H g per H22O moleculeO molecule

• What is the mass of 12.0 million benzene (CWhat is the mass of 12.0 million benzene (C66HH66) )

molecules?molecules? Molar mass = (6 mol C)(12.01 g/mol C) Molar mass = (6 mol C)(12.01 g/mol C)

+ (6 mol H)(1.008 g/mol H) = 78.11 g/mol C + (6 mol H)(1.008 g/mol H) = 78.11 g/mol C66HH66

12.0 million= 12,000,000 = 1.20 x 1012.0 million= 12,000,000 = 1.20 x 1077 molecules molecules

1.20 x 107 molec.x 1 mol x 78.11 g 6.022 x 1023 molecules 1 mol

78.11 g78.11 g= 1.99 x 10= 1.99 x 10–17 –17 mol Cmol C66HH6 6 ────────────= = mol Cmol C66HH66

1.56 x 101.56 x 10–15–15 g CC66HH66

Consider 4.49 g CaConsider 4.49 g Ca33(PO4)2 . Its formula weight FW = 3x40.08 + 2x(30.97 + 4x16.00)=310.18 g/mol

• How many mol of CaHow many mol of Ca33(PO4)2 is this? is this? 1 mol 1 mol 4.49 g 4.49 g x x ────────── = = 0.01450.0145 moles Ca moles Ca33(PO4)2 310.18 g310.18 g

• How many mol of Ca2+ ions does this contain? 3 moles Ca3 moles Ca2+2+ 0.0145 mol 0.0145 mol CaCa33(PO4)2 x x────────────── = = 0.04350.0435 mol Ca mol Ca2+2+

1 mol CaCa33(PO4)2

• How many Ca2+ ions are there? 6.022 x1023 Ca Ca2+2+ 0.0435 mol 0.0435 mol CaCa2+2+ x x ─────────── ───────────= 2.61x10= 2.61x102222 ions ions 1 mol CaCa2+2+

Consider 4.49 g CaConsider 4.49 g Ca33(PO(PO44))22

FW = 310.18 g/molFW = 310.18 g/mol• How many mol of P does that amount of CaHow many mol of P does that amount of Ca33(PO4)2

contain?contain? 2 moles P 2 moles P 0.0145 mol 0.0145 mol CaCa33(PO4)2 x x────────────── = 0.0290 mol P = 0.0290 mol P 1 mol CaCa33(PO4)2

• How many mol of O does that amount contain?How many mol of O does that amount contain? 8 moles O 8 moles O 0.0145 mol 0.0145 mol CaCa33(PO4)2 x x────────────── = 0.116 mol O = 0.116 mol O 1 mol CaCa33(PO4)2

• How many grams of O are in that amount of salt?How many grams of O are in that amount of salt? 16.0 g O 16.0 g O 0.116 mol O 0.116 mol O xx────────────── = 1.86 g O = 1.86 g O 1 mol O

Consider 4.49 g CaConsider 4.49 g Ca33(PO(PO44))22

FW = 310.18 g/molFW = 310.18 g/molHow many phosphate ions does that amount of How many phosphate ions does that amount of CaCa33(PO4)2 contain? contain?

First we calculate how many moles of phosphateFirst we calculate how many moles of phosphate

2 moles PO2 moles PO4433−−

0.0145 mol 0.0145 mol CaCa33(PO4)2 x x────────────────── = 0.0290 mol PO = 0.0290 mol PO4433−−

1 mol CaCa33(PO4)2

Second, using Avogadro’s number,

6.02 x 106.02 x 102323 PO PO4433−−

0.0290 mol 0.0290 mol POPO4433−− x x ────────────────────── = 1.75 x 10 = 1.75 x 102222 POPO44

33−− 1 mol POPO44

33−− ionsions

What mass of CWhat mass of C44HH88O is required to O is required to

supply 1.43 x 10supply 1.43 x 101616 C atoms? C atoms?

• Molar mass = 72.10 g/molMolar mass = 72.10 g/mol

1 mol C atoms 1 mol C1 mol C atoms 1 mol C44HH88OO1.43 x 101.43 x 101616 C atoms C atoms x x──────────── x ─────────────────── x ─────── 6.022x10.022x1023 23 C atoms 4 moles C C atoms 4 moles C 72.10 g C72.10 g C44HH88OOxx──────────── = 4.28 x 10──────────── = 4.28 x 10−7−7 g g CC44HH88OO 1 mol C1 mol C44HH88OO

Empirical Formula from AnalysisEmpirical Formula from AnalysisIt will be shown by example.It will be shown by example.

We will use We will use The Percent Composition, i.e, The Percent Composition, i.e, the amount (grams) of each element the amount (grams) of each element

forming one compound in 100 g of that forming one compound in 100 g of that compound.compound.

((An example of calculating %sAn example of calculating %s))

For the compound AFor the compound AxxBByyCCzz ( (x, y, z are x, y, z are

unknownunknown))

% A + % B + % A + % B + %%C = 100C = 100

• A compound was found to be 40.92 A compound was found to be 40.92 %% C, C, 4.58 4.58 %% H, and 54.50 H, and 54.50 %% O by mass. O by mass.

• Determine its empirical formula.Determine its empirical formula.

CCxxHyOz

x, y, and z are the simplest whole number ratios of atoms in compound.

Assume exactly 100 g of compoundthen 40.92 g C, 4.58 g H, and 54.50 g O

are contained in those 100 g

Firstly, calculate the number of moles of C, H, and O in that amount of compound… next

1 mole C1 mole C40.92 g C x40.92 g C x──────── = 3.407 moles C──────── = 3.407 moles C 12.01 g C12.01 g C

1 mole H1 mole H4.58 g H x 4.58 g H x ──────── = 4.54 moles H──────── = 4.54 moles H 1.008 g H1.008 g H

1 mole O1 mole O54.50 g O x54.50 g O x──────── = 3.406 moles O──────── = 3.406 moles O 16.00 g O16.00 g O

CC3.4073.407HH4.544.54OO3.4063.406 Now, divide all # by the smallest Now, divide all # by the smallest

CC3.4073.407HH4.544.54OO3.4063.406 The result is C The result is C1.01.0HH1.331.33OO1.01.0 3.406 3.406 3.406

Finally, 1.33 x 3 = 4 Finally, 1.33 x 3 = 4 Empirical formula isEmpirical formula is

And the same for 1.0 x 3 And the same for 1.0 x 3 CC33HH44OO33

Determine the empirical formula of a compound Determine the empirical formula of a compound that is 36.4% Mn, 21.2% S, and 42.4% O by mass.that is 36.4% Mn, 21.2% S, and 42.4% O by mass.

assuming exactly 100 g compoundassuming exactly 100 g compound 1 mole Mn1 mole Mn36.4 g Mn x36.4 g Mn x──────── = 0.663 moles Mn──────── = 0.663 moles Mn 54.9 g Mn54.9 g Mn

1 mole S1 mole S21.2 g S x 21.2 g S x ──────── = 0.660 moles S──────── = 0.660 moles S 32.1 g S32.1 g S

1 mole O1 mole O42.4 g O x42.4 g O x──────── = 2.65 moles O──────── = 2.65 moles O 16.0 g O16.0 g O

MnMn0.6630.663SS0.6600.660OO2.652.65 The result is Mn The result is Mn1.01.0SS1.01.0OO4.024.02 0.660 0.660 0.660

empirical formula: MnSOempirical formula: MnSO44

Homework: Determine the empirical Homework: Determine the empirical formula of a compound that is formula of a compound that is 27.6% Mn, 24.2% S, & 48.2% O27.6% Mn, 24.2% S, & 48.2% O

answeranswer

MnMn22SS33OO1212

molecular formula = n x (empirical formula)molecular formula = n x (empirical formula)

• n = whole numbern = whole number molar mass (or formula weight)molar mass (or formula weight)• n = n = ────────────────────────────────────────────────── weight of the empirical formulaweight of the empirical formula

Example: benzene, molecular formula: CExample: benzene, molecular formula: C66HH66

empirical formula: CHempirical formula: CH

n = 6 6 x (CH) = n = 6 6 x (CH) = CC66HH66

Example: A 4.99 g sample of a compound was Example: A 4.99 g sample of a compound was found to contain 1.52 g N and oxygen. found to contain 1.52 g N and oxygen. Determine its empirical and molecular formula Determine its empirical and molecular formula if its MW is 92.04if its MW is 92.04• determine empirical formula g O = 4.99 −1.52 1 mol N = 3.47 g1.52 g N─────── = 0.108 mol N 14.01 g N

1 mol O3.47 g O ─────── = = 0.217 mol O 16.00 g O

NN0.1080.108OO0.2170.217 NO NO22 0.108 0.108

• determine molecular formuladetermine molecular formula empirical formula NOempirical formula NO22

empirical weight = 14.01+ 2empirical weight = 14.01+ 216.00 = 46.0116.00 = 46.01 molecular weight = 92.04 (molecular weight = 92.04 (given in given in

previous slideprevious slide)) 92.0492.04 n = n = ────────── = 2 = 2 46.0146.01 Molecular formula = 2 x (emp. form.)Molecular formula = 2 x (emp. form.) = 2(NO= 2(NO22))

molecular formula = Nmolecular formula = N22OO44

A compound was found to contain only B A compound was found to contain only B and H. Analysis of an 8.247 g sample and H. Analysis of an 8.247 g sample indicated that it contained 1.803 g H and indicated that it contained 1.803 g H and had a MW of ~30. Determine its molecular had a MW of ~30. Determine its molecular formula and its MW to 4 SF.formula and its MW to 4 SF.

• determine empirical formuladetermine empirical formula

total mass = g B + g Htotal mass = g B + g H

8.247 g = g B + 1.803 g H8.247 g = g B + 1.803 g H

g B = 8.247 g compound g B = 8.247 g compound −−1.803 g H =6.444 g 1.803 g H =6.444 g

1 mol B1 mol B6.444 g B 6.444 g B ───────= 0.5961 mol of boron───────= 0.5961 mol of boron 10.811 g B10.811 g B

1 mol H1 mol H 1.803 g H 1.803 g H ────────────── = 1.789 mol H = 1.789 mol H 1.008 g H1.008 g H

BB0.59610.5961 H H1.7891.789 BH BH33 is the Empirical formula is the Empirical formula

0.59610.5961 0.59610.5961

• determine molecular formuladetermine molecular formula MW 30MW 30 n = n = ────────────── ──────── = 2 = 2 emp. wt. 13.83emp. wt. 13.83

mol. form. = 2(BHmol. form. = 2(BH33) = B) = B22HH66

BB22HH66 MW = 2 MW = 210.811 + 610.811 + 6 1.008 1.008 = 27.67 g/mol

HydratesHydrates compounds that contain intact Hcompounds that contain intact H22OO

eg. CaSOeg. CaSO44•2H•2H22OO

Two HTwo H22O moles per mole of CaSOO moles per mole of CaSO44

compound name compound name prefixprefix-hydrate-hydrate

(prefix(prefix indicates number of H indicates number of H22O molecules)O molecules)

calcium sulfate calcium sulfate didihydratehydrate

• LiCl•8HLiCl•8H22OO

lithium chloride lithium chloride octaoctahydratehydrate

• tungsten(VI) sulfate tungsten(VI) sulfate heptaheptahydratehydrate

WW6+6+

SOSO442–2–

W(SOW(SO44))33•7H•7H22OO

• Hydrated nickel(II) chloride is a beautiful green color, Hydrated nickel(II) chloride is a beautiful green color, crystalline compound. When heated strongly, the crystalline compound. When heated strongly, the compound is dehydrated. If 0.235 g of NiClcompound is dehydrated. If 0.235 g of NiCl22•xH•xH22O gives O gives 0.128 g of 0.128 g of NiClNiCl22 on heating, what is the value of x? on heating, what is the value of x?

g of NiClg of NiCl22•xH•xH22O = g O = g NiClNiCl22 + g H + g H22O Then,O Then,

hydrated dehydratedhydrated dehydrated

g Hg H22O = 0.235 g – 0.128 g = 0.107 gO = 0.235 g – 0.128 g = 0.107 g

1 mol NiCl1 mol NiCl22

0.128 g 0.128 g NiClNiCl22 x ──────── = 0.000987 moles x ──────── = 0.000987 moles NiClNiCl22

129.7 g NiCl129.7 g NiCl22

1 mol H1 mol H22OO

0.107 g H0.107 g H22O x ──────── = 0.00594 moles HO x ──────── = 0.00594 moles H22OO

18.0 g 18.0 g HH22OO

NiClNiCl22 0.0009870.000987 H H22OO0.005940.00594 NiClNiCl22•6H•6H22O x = 6 moles of waterO x = 6 moles of water

0.0009870.000987 0.000987 0.000987 per mole of NiClper mole of NiCl22

Percent CompositionPercent Composition

The percent of the total mass of a The percent of the total mass of a substance represented by each element substance represented by each element

within that substancewithin that substance

In other words, how many grams of every In other words, how many grams of every element are in 100 g of the compoundelement are in 100 g of the compound

Determine the % composition of aluminum Determine the % composition of aluminum bromitebromite

• AlAl3+3+

• BrO2–

• Al(BrO2)3 Al Br O

• FW = 26.98226.982 + 3 x(79.904 + 2 x 16.000)

or = 26.982 + 3x79.904 + 6x16.000

= 362.68 g/mol

g Alg Al % Al = ──────── x 100% Al = ──────── x 100 g compoundg compound

26.982 g 26.982 g % Al = ──────── x 100 = 7.440 %% Al = ──────── x 100 = 7.440 % 362.68 g 362.68 g

g Brg Br % Br = ──────── x 100% Br = ──────── x 100 g compoundg compound 3 x 79.904 g 3 x 79.904 g % Br = ──────── x 100 = 66.095 %% Br = ──────── x 100 = 66.095 % 362.68 g 362.68 g

% O = % O = 100% – 7.440% – 66.095 = 26.465% 100% – 7.440% – 66.095 = 26.465%

3.69 Silver chloride, often used in silver plating, 3.69 Silver chloride, often used in silver plating, contains 75.27% Ag. What mass of silver chloride is contains 75.27% Ag. What mass of silver chloride is required to plate 155 mg of pure silver? AgClrequired to plate 155 mg of pure silver? AgCl

The 75.27% means there is 75.27 g of Ag in 100 g AgCl.The 75.27% means there is 75.27 g of Ag in 100 g AgCl.That can be used as a conversion factor (ratio) That can be used as a conversion factor (ratio)

111010−3−3 g Ag g Ag 155 mg Ag 155 mg Ag ──────── = 0.155 g Ag ──────── = 0.155 g Ag 1 mg1 mg 100100 g AgCl g AgCl0.155 0.155 g Agg Ag ──────── = 0.206 g AgCl ──────── = 0.206 g AgCl 75.27 75.27 g Ag g Ag or, with mg,or, with mg, 100100 mg AgCl mg AgCl155 155 mg Agmg Ag ──────── = 206 mg AgCl ( ──────── = 206 mg AgCl (with mgwith mg)) 75.27 75.27 mg Agmg Ag